In this section, we will generalize the notion of orthogonality of vectors in \mathbb{R}^n from two vectors to sets of vectors. In doing so we will see that two properties make the standard basis \{ e_1,e_2,...,e_n \} of \mathbb{R}^n easy to work with: First, any two distinct vectors in the set are orthogonal. Second, each vector in the set is a unit vector. These two properties lead us to the notion of orthogonal bases and orthonormal bases--concepts that we will be able to apply to various applications.
A set of vectors \{\vec{v}_1,\vec{v}_2,...,\vec{v}_k \} in \mathbb{R}^n is called an orthogonal set if all pairs of distinct vectors in the set are orthogonal--that is, if \vec{v}_i \cdot \vec{v}_j = 0 whenever j \neq i for i,j = 1,2,...,k.
Example: Show that \vec{v}_1 = \left[ \begin{array}{c} 2 \\ 1 \\ -1 \end{array} \right], \vec{v}_2 = \left[ \begin{array}{c} 0 \\ 1\\ 1\end{array}\right], \vec{v}_3 = \left[ \begin{array}{c} 1 \\ -1 \\ 1 \end{array} \right] form an orthogonal set in \mathbb{R}^3.
One of the main advantages to working with orthogonal sets of vectors is that they are necessarily linearly independent:
If \{ \vec{v}_1,\vec{v}_2,...,\vec{v}_k\} is an orthogonal set of nonzero vectors in \mathbb{R}^n, then these vectors are linearly independent.
Remark: Thanks to this theorem, we know that if a set of nonzero vectors in orthogonal, they are automatically linearly independent. Thus the vectors \vec{v}_1,\vec{v}_2,\vec{v}_3 are automatically linearly independent!
An orthogonal basis for a subspace W of \mathbb{R}^n is a basis of W that is an orthogonal set.
So \{\vec{v}_1,\vec{v}_2,\vec{v}_3 \} is an orthogonal basis for \mathbb{R}^3.
Example: Find an orthogonal basis for the subspace W of \mathbb{R}^3 given by
W = \{ \left[ \begin{array}{c} x \\ y \\ z\end{array} \right] : x +2y -z = 0 \}.
Solution: We have x = -2y + z so any vector in W is of the form \left[ \begin{array}{c}-2y +z \\ y \\ z \end{array} \right] = y \left[ \begin{array}{c} -2 \\ 1\\ 0 \end{array} \right] + z \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right].
It follows that \vec{u} = \left[ \begin{array}{c} -2 \\ 1\\ 0 \end{array} \right] and \vec{v}= \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right] form a basis for W.
But \vec{u} \cdot \vec{v} = -2 \neq 0, so \{ \vec{u},\vec{v} \} is not an orthogonal basis for W.
To find an orthogonal basis for W, is suffices to find a nonzero vector in W that is orthogonal to either one of these. So we'll find a vector orthogonal to \vec{u}.
Suppose \vec{w} = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] is a vector in W that is orthogonal to \vec{u}.
Then x+2y -z= 0, since \vec{w} \in W and \vec{u} \cdot \vec{w} = -2x+y = 0.
Solving these two linear equations, we find y = 2x and z = 5x or \vec{w} = x \left[ \begin{array}{c} 1 \\ 2 \\ 5\end{array} \right].
To be specific, we could take \vec{w} = \left[ \begin{array}{c} 1 \\ 2\\ 5\end{array} \right].
It is easy to check that \{ \vec{u}, \vec{w} \} is an orthogonal set in W and, hence, and orthogonal basis for W since dim W = 2.
Exercise: Find an orthogonal basis containing \vec{v}.
Remark: Another advantage of working with an orthogonal basis is that the coordinates of a vector with respect to such a basis are easy to compute. Indeed, there is a formula for these coordinates, as the following theorem establishes.
Let \{ \vec{v}_1,\vec{v}_2,...,\vec{v}_k\} be an orthogonal basis for a subspace W of \mathbb{R}^n and let \vec{w} be any vector in W.
Then the unique scalars c_1,...,c_k, such that
\vec{w} = c_1 \vec{v}_1 + \cdots +c_k \vec{v}_k
are given by
c_i = \frac{\vec{w} \cdot \vec{v}_i}{\vec{v}_i \cdot \vec{v}_i} for i = 1,...,k.
Exercise: Find the coordinates of \vec{w} = \left[\begin{array}{c} 1 \\ 2 \\3 \end{array} \right] with respect to the orthogonal basis \{\vec{v}_1,\vec{v}_2,\vec{v}_3\}.
Notice that yet again, if we have orthogonal vectors, we no longer need to row reduce.
A set of vectors in \mathbb{R}^n is an orthonormal set if it is an orthogonal set of unit vectors.
An orthonormal basis for a subspace W of \mathbb{R}^n is a basis of W that is an orthonormal set.
If S = \{ \vec{q}_1,...,\vec{q}_k\} is an orthonormal set of vectors, then \vec{q}_i \cdot \vec{q}_j = 0 for j \neq i and ||\vec{q}_i||= 1.
The fact that each \vec{q}_i is a unit vector is equivalent to \vec{q}_i\cdot \vec{q}_i = 1.
If follows that we can cummarize the statement that S is orthonormal as
\vec{q}_i \cdot \vec{q}_i = \begin{array}{rr} 0 & j \neq i \\ 1 & j = i \end{array}
Exercise: Is the orthogonal basis \{\vec{v}_1,\vec{v}_2,\vec{v}_3\} an orthonormal basis?
How can you make it into an orthonormal basis?
What is this new basis?
Let \{ \vec{q}_1,...\vec{q}_k\} be an orthonormal basis for a subspace W of \mathbb{R}^n and let \vec{w} be any vector in W.
Then \vec{w} = (w \cdot \vec{q}_1)\vec{q}_1+\cdots+(w\cdot \vec{q}_k) \vec{q}_k.
and this representation is unique.
Matrices whose columns form an orthonormal set arise frequently in applications. Such matrices have several nice properties, which we will now examine.
The columns of an m \times n matrices Q form an orthonormal set if and only if Q^T Q = I.
If this matrix Q is a square matrix, it has a special name:
An n \times n matrix Q whose columns form an orthonormal set is called an orthogonal matrix.
The most important fact about orthogonal matrices is given by the next theorem.
A square matrix Q is orthogonal if and only if Q^{-1} = Q^T.
Examples: Show that the following matrices are orthogonal and find their inverses:
A = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]
B = \left[ \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right]
Remark: Matrix A is an example of a permutation matrix, a matrix obtained by permuting the columns of an identity matrix. In general, any n \times n permutation matrix is orthogonal. Matrix B is the matrix of a rotation through the angle \theta in \mathbb{R}^2. Any rotation has the property that it is a length-preserving transformation (known as an isometry in geometry). The next theorem shows that every orthogonal matrix transformation is an isometry. Orthogonal matrices also preserve dot products. In fact, orthogonal matrices are characterized by either one of these properties.
Let Q be an n\times n matrix. The following statements are equivalent:
a. Q is orthogonal
b. ||Q \vec{x}|| = || \vec{x}|| for every \vec{x} in \mathbb{R}^n
c. Q \vec{x} \cdot Q \vec{y} = \vec{x} \cdot \vec{y} for every \vec{x} and \vec{y} in \mathbb{R}^n.
Not only do the columns of orthogonal matrices form orthonormal sets -- so do the rows:
If Q is an orthogonal matrix, then its rows form an orthonormal set.
Here are some other awesome properties:
Let Q be an orthogonal matrix.
a. Q^{-1} is orthogonal
b. det Q = \pm 1
c. If \lambda is an eigenvalue of Q, then |\lambda| = 1 (here | \cdot | represents the absolute value)
d. If Q_1 and Q_2 are orthogonal n \times n matrices, then so is Q_1 Q_2.
Remark: Property (c) holds even for complex eigenvalues.
|
|